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The given limit problem is:

\[
\lim_{x \to 1} \frac{4x + 3}{x - 1}
\]

### Solution:
1. **Substitute \( x = 1 \) directly:**

\[
\frac{4(1) + 3}{1 - 1} = \frac{4 + 3}{0} = \frac{7}{0}
\]

Since the denominator becomes zero, this limit indicates a vertical asymptote, implying that the expression tends towards infinity or negative infinity. To determine this precisely, we need to analyze the behavior of the function as \( x \to 1^+ \) (from the right) and \( x \to 1^- \) (from the left).

2. **As \( x \to 1^+ \) (from the right):**

For values of \( x \) slightly greater than 1, \( x - 1 \) is positive. Therefore,

\[
\frac{4x + 3}{x - 1} \to +\infty
\]

3. **As \( x \to 1^- \) (from the left):**

For values of \( x \) slightly less than 1, \( x - 1 \) is negative. Therefore,

\[
\frac{4x + 3}{x - 1} \to -\infty
\]

### Conclusion:
The limit does not exist as it approaches different infinities from the left and right:

\[
\lim_{x \to 1} \frac{4x + 3}{x - 1} = \text{DNE} \quad (\text{Does Not Exist}).
\]

Would you like any further details or clarifications?

Here are 5 related questions for further exploration:
1. What is the general approach to solving limits involving division by zero?
2. How do you determine if a limit approaches \( +\infty \) or \( -\infty \)?
3. Can this limit be evaluated using L'Hopital's Rule?
4. What is a vertical asymptote, and how is it related to limits?
5. How does one interpret one-sided limits in real-world scenarios?

**Tip:** When dealing with limits that result in a zero denominator, always check the behavior from both the left and right sides of the point of interest.

Understand how to solve the limit lim_{x -> 1} (4x + 3) / (x - 1) by analyzing its one-sided behavior. The function exhibits a vertical asymptote as x approaches 1. Explore key concepts in limits and asymptotes, suitable for advanced high school students.

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